A particle executes SHM such that at a given time it is at x= +1/4 the amplitude, moving away from equilibrium, and 0.7 seconds later the particle has -1/6 the maximum speed moving away from equilibrium. Find the period of the motion.
To find the period of the motion in this problem, we can use the given information about the particle's position and velocity.
Let's break down the information given in the problem:
Given:
- At a specific time, the particle is at x = +1/4 the amplitude, moving away from equilibrium.
- 0.7 seconds later, the particle has -1/6 the maximum speed, still moving away from equilibrium.
First, let's define some variables:
- A: Amplitude of the motion
- t1: Initial time (when the particle is at x = +1/4 the amplitude)
- t2: Time after 0.7 seconds (when the particle has -1/6 the maximum speed)
The first piece of information tells us that at t = t1, the particle is at x = +1/4A. This means that the displacement from the equilibrium position can be written as x(t1) = (1/4)A.
The second piece of information tells us that at t = t2 (0.7 seconds later), the particle has -1/6 the maximum speed. The velocity of a particle in simple harmonic motion can be written as v(t) = (dx/dt), where x(t) represents the displacement of the particle. Since the velocity is negative and moving away from equilibrium, we have v(t2) = -(1/6)v_max.
Now, let's use some equations for simple harmonic motion:
1) The position of the particle as a function of time is given by x(t) = A * cos[(2π/T)t + φ], where T is the period and φ is the phase constant.
2) The velocity of the particle as a function of time is given by v(t) = -(2π/T)A * sin[(2π/T)t + φ].
Using equation 1, we can substitute t = t1 and x(t1) = (1/4)A to find the phase constant φ.
(1/4)A = A * cos[(2π/T)t1 + φ]
Dividing both sides by A:
(1/4) = cos[(2π/T)t1 + φ]
Since cosine is a periodic function, we know that cos(θ) = cos(θ + 2πn), where n is an integer. Therefore, we can assume that:
(2π/T)t1 + φ = 2πn
Simplifying:
(2π/T)t1 = 2πn - φ
Now, let's use equation 2 to find the value of φ.
-(1/6) = -(2π/T)A * sin[(2π/T)t2 + φ]
Dividing both sides by -(2π/T)A:
(1/6) = sin[(2π/T)t2 + φ]
Similar to before, we can assume that:
(2π/T)t2 + φ = 2πn
Simplifying:
(2π/T)t2 = 2πn - φ
We can set these two expressions equal to each other:
(2π/T)t1 = (2π/T)t2
Simplifying and canceling out the 2π terms:
t1/T = t2/T
Now, we have the relationship t1 = t2, which means the time intervals at which the particle has the given position and velocity are equal. Therefore, we can say t1 = 0.7 seconds.
Now, we can solve for the period T:
t1/T = 0.7/T
Simplifying:
1/T = 0.7
T = 1/0.7
T ≈ 1.428 seconds
So, the period of the motion is approximately 1.428 seconds.